A is diagonalizable iff there are n linearly independent eigenvectors

Dependencies:

  1. Diagonalization
  2. Linear independence
  3. Inverse of a matrix
  4. Transpose of product
  5. Full-rank square matrix is invertible
  6. A matrix is full-rank iff its rows are linearly independent

Let $A$ be an $n$ by $n$ matrix. Then $L$ is diagonalizable iff there are $n$ linearly independent eigenvectors for $A$.

Proof

\begin{align} & P^{-1} \textrm{ exists} \\ &\iff \exists Q, QP = PQ = I \\ &\iff \exists Q, P^TQ^T = Q^TP^T = I \\ &\iff (P^T)^{-1} \textrm{ exists} \\ &\iff \operatorname{rank}(P^T) = n \\ &\iff \textrm{Rows of } P^T \textrm{ are linearly independent} \\ &\iff \textrm{Columns of } P \textrm{ are linearly independent} \end{align}

$(\exists P, \exists \textrm{ diagonal } D, AP = PD) \iff$ columns of $P$ are eigenvectors of $A$.

If there are $n$ linearly independent eigenvectors, make them the columns of $P$. Then $AP = PD$ ($D$ is diagonal) and $P^{-1}$ exists, so $D = P^{-1}AP$. Therefore, $A$ is diagonalizable.

If $A$ is diagonalizable, there is a $P$ such that $P^{-1}$ exists and $AP = PD$ ($D$ is diagonal). Therefore, columns of $P$ are linearly independent and they are eigenvectors of $A$. Therefore, $A$ has $n$ linearly independent eigenvectors.

Dependency for: None

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Transitive dependencies:

  1. /linear-algebra/matrices/gauss-jordan-algo
  2. /linear-algebra/vector-spaces/condition-for-subspace
  3. /sets-and-relations/equivalence-relation
  4. /sets-and-relations/composition-of-bijections-is-a-bijection
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  25. Reduced Row Echelon Form (RREF)
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  27. Elementary row operation
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  31. Row space
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  39. Row equivalence matrix
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  53. Canonical decomposition of a linear transformation
  54. Eigenvalues and Eigenvectors
  55. Diagonalization
  56. A matrix is full-rank iff its rows are linearly independent
  57. Full-rank square matrix is invertible